如何获取列表中第一个元素中最小值的所有元组对

Question

输出只是列表中的第一个实例，而我想要 [(1,'a'),(1,'b')] 的两个实例

smallest_teams = []

roster = [(1,'a'),(1,'b'),(3,'c')]

smallest_teams = list(min(roster))
print(smallest_teams)

输出只是列表中的第一个实例，而我想要 [(1,'a'),(1,'b')] 的两个实例

smallest_teams = []

roster = [(1,'a'),(1,'b'),(3,'c')]

smallest_teams = list(min(roster))
print(smallest_teams)

Answer 1

您可以使用 for 循环实现您想要的结果。您可以随时跟踪列表中最小的团队，并在遇到较小的值时立即重置列表

roster = [(1, 'a'), (1, 'b'), (3, 'c')]
min_value = float('inf')  # Initialize min_value to positive infinity
smallest_teams = []

for item in roster:
    if item[0] < min_value:
        min_value = item[0]
        smallest_teams = [item]
    elif item[0] == min_value:
        smallest_teams.append(item)

print(smallest_teams)